Simplify two expressions and draw circuits

i) A'B'C' + A'B'C + AB'C' + AB'C

Group terms with B' common:

= B'(A'C' + A'C + AC' + AC)
= B'[A'(C' + C) + A(C' + C)]
= B'[A' + A]
= B'

Answer: F = B'

Circuit: only one NOT gate (inverter) with input B.

 B ---[NOT]--- F = B'

ii) AC' + ABC + A(C + AC')

A(C + AC') = AC + AAC' = AC + AC'
So: AC' + ABC + AC + AC'
  = AC' + ABC + AC            (AC' + AC' = AC')
  = A(C' + C) + ABC
  = A + ABC
  = A(1 + BC)
  = A

Answer: F = A

Circuit: a direct connection, with no gate. The output follows A.

 A ----------------- F = A

Both circuits show how Boolean algebra removes gates: the original needed several AND, OR and NOT gates, but the simplified forms need one NOT gate and a plain wire.